probability question and any theorem related to it. A man speaks true $3/4$ times. A die is thrown and he reports it to be $4$ . what is the probability that it is actually four. 
MY TRY
So probability of him speaking truth is $3/4$ so favourable cases are $3/4 \times 1/6=1/8$ now total cases are where he speaks true, false so its $3/4 \times 1/6+1/4 \times 5/6=1/3$ so probability is $3/8$ . Is my reasoning right? And is there any theorem for these sums.
 A: Perhaps it's just the phrasing, but I would not agree with your calculation.  
The problem does not explain the nature of a lie. I would have thought the natural interpretation here was that when he lies he chooses a false answer uniformly among the options.  Assuming this to be the correct reading your second term ought to read $$\frac 14\times\frac 56\times \frac 15=\frac 1{24}$$  
Further, your "denominator" needs to be the total probability that he says "I threw a $4$".  Thus I would see the correct answer as $$\frac {\frac 16\times \frac 34}{\frac16\times \frac 34+\frac 1{24}}=\frac {\frac 18}{\frac 4{24}}=\frac 34$$
A: It is given that the man says $4$
Assuming that when he lies, he chooses the number randomly,
P(actually $4$ | man says $4$) = $\dfrac {(1/6 \times 3/4)}{(1/6\times 3/4) + (5/6\times 1/5\times 1/4)}= 3/4$
Added break-up of denominator and reference to another question
P($4$ and says $4$) = $1/6\times 3/4$
P(not $4$ and says $4$) = $5/6\times 1/5\times 1/4$
See a more complicated version of a similar question here
